삼각함수 극한 도형

[출제의도] 삼각함수의 극한을 활용하여 문제해결하기 $\displaystyle \mathrm{\overline{AD}} = \overline{CD}$이므로 삼각형 $\mathrm{ADC}$는 이등변삼각형이고, $\displaystyle \mathrm{\overline{AE}} = {\overline{CE}}$이다. 또한, 선분 $\mathrm{AC}$와 선분 $\mathrm{DF}$가 평행하므로 $\angle \mathrm{BDF} = \theta$이고, 삼각형 $\mathrm{DEF}$는 직각삼각형이다.

삼각형 $\mathrm{ABC}$에서 $\displaystyle {\overline{\mathrm{AC}}} \mathit{=} \sec \theta$이므로 $\displaystyle {\overline{\mathrm{AE}}} \mathit{=} \frac{1}{2} {\mathrm{\sec}} \theta$ 삼각형 $\mathrm{AED}$에서 $\displaystyle {\overline{\mathrm{DE}}} \mathit{=} {\overline{\mathrm{AE}}} \times \tan \theta = \frac{1}{2} \mathrm{\sec} \theta \tan \theta$, $\displaystyle {\overline{\mathrm{AD}}} \mathit{=} {\overline{\mathrm{AE}}} \times \sec \theta = \frac{1}{2} \mathrm{\sec} ^{2} \theta$이므로 $\displaystyle {\overline{\mathrm{DB}}} \mathit{=} 1 - {\frac{1}{2}} \mathrm{\sec} ^{2} \theta$ 삼각형 $\mathrm{DBF}$에서 $\displaystyle {\overline{\mathrm{DF}}} \mathit{=} \overline{\mathrm{DB}} \times \sec \theta = \sec \theta \left( 1 - \frac{1}{2} \mathrm{\sec} ^{2} \theta \right)$ $S \left( \theta \right)$$\displaystyle = \frac{1}{2} \times \mathrm{\overline{DE}} \times {\overline{DF}}$ $\displaystyle = \frac{1}{2} \times \frac{1}{2} \mathrm{\sec} \theta \tan \theta \times \sec \theta \left( 1 - \frac{1}{2} \mathrm{\sec} ^{2} \theta \right)$ $\displaystyle = \frac{1}{4} \mathrm{\sec} ^{2} \theta \tan \theta \left( 1 - \frac{1}{2} \mathrm{\sec} ^{2} \theta \right)$ $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {\frac{S \left( \theta \right)}{\theta}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {\frac{\frac{1}{4} \mathrm{\sec} ^{2} \theta \tan \theta \left( 1 - \frac{1}{2} \mathrm{\sec} ^{2} \theta \right)}{\theta}}$ $\displaystyle = \frac{1}{8}$